Calculate $$\lim\limits_{n\to\infty}(1-\frac{\lambda_n}n)^n$$

I came upon the calculation of $\lim\limits_{n\to\infty}(1-\frac{\lambda_n}n)^n$ when I was reviewing Poisson distribution. Notice that $\lambda_n = np_n(n\in\mathbb{Z}_{\geq 0}, 0\leq p \leq 1)$ and $\lim\limits_{n\to\infty}\lambda_n = \lambda(\lambda\in\mathbb{R})$.

So how to calculate it? The first thought that came to my mind was that

$$\lim\limits_{n\to\infty}(1 + \frac1 n)^n = e \tag 1$$

So replace $n$ with $-n$, I got

$$\lim\limits_{n\to\infty}(1 - \frac1 n)^n = \frac1 e \tag 2$$

Then I have no clue about what to do next, I’m not even sure whether (2) is right or not. So I started digging from the calculation of $\lim\limits_{n\to\infty}(1 + \frac1 n)^n$.

Here is how I solve the problem.

Let $t$ be any number in the interval $[1-\frac{\lambda_n}n, 1]$, then we get

Taking the $(\lambda_n - n)^{th}$ power of the right inequality gives us

$$(1-\frac{\lambda_n} n)^{(n-\lambda_n)}\geq e^{-\lambda_n}$$

Why was $\geq$ replaced with $\leq$? Because $\lambda_n = np_n, n\geq 0, 0\leq p_n\leq 1$($p_n$ denotes the probability), so $\lambda_n-n = np_n - n = n(p_n - 1)\leq 0$, the power is less or equal to 0, so we need change the direction of the sign.

Multiply each side of the inequality by $(1-\frac{\lambda_n} n)^{\lambda_n}$, we get